User:Egm6321.f12.team4.mishra/Homework4 R4.5

Homework 4 Problem R*4.5 – Symmetry of mixed partial derivatives
sec22-4

Statement
Using the general form of the 2nd exactness condition for Nn-ODE, show the equivalence to symmetry of mixed 2nd partial derivatives of the first integral $$ \phi $$

Solution
We know that $$ \phi $$ is a function of x,y,y'.

Taking the total derivative of $$ \phi $$ w.r.t x.


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$$ \frac{d}{dx}\phi= \phi_x + \phi_y y' + \phi_{y'} y'' $$ (4.5.1)
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Since,


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$$ g_i = \frac{\partial }{\partial y^{i}}(\frac{d}{dx}\phi) $$ (4.5.2)
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$$ g_0= \frac{\partial}{\partial y} (\frac{d\phi}{dx}) = \phi_{xy}+\phi_{yy}y'+\phi_{yy'}y'' $$ (4.5.3)
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$$ g_1= \frac{\partial}{\partial y'} (\frac{d\phi}{dx}) = \phi_{xy'}+\phi_{yy'}y'+\phi_y+ \phi_{y'y'}y'' $$ (4.5.4)
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$$ g_2= \frac{\partial}{\partial y''} (\frac{d\phi}{dx}) = \phi_y' $$ (4.5.5)
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For the exactness condition,


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$$ g_0 + \frac{d}{dx} g_1 + \frac{d^2}{dx^2} g_2 = 0 $$ (4.5.6)
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Differentiating  with respect to x once,


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$$ \frac{d}{dx} g_2 = \phi_{y'x} + \phi_{y'y}y'+ \phi_{y'y'}y'' $$ (4.5.7)
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Putting in the values of $$ g_0, g_1, g_2 $$ from Eqns. (4.5.2 - 4.5.4) and 4.5.7,


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$$ g_0 - \frac{d}{dx} g_1 + \frac{d^2}{dx^2} g_2 = 0 $$ (4.5.7)
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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} (\phi_{xy'}+\phi_{yy'}y'+\phi_y+ \phi_{y'y'}y) + \frac{d}{dx} (\phi_{y'x} + \phi_{y'y}y'+ \phi_{y'y'}y) =0 $$ (4.5.8)
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The above can be rearranged to be written as


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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} \phi_y - \frac{d}{dx} (\phi_{xy'}-\phi_{y'x}+\phi_{yy'}y'-\phi_{y'y}y') =0 $$ (4.5.9)
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Considering the part outside the derivative,


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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} \phi_y = (\phi_{xy}+\phi_{yy}y'+\phi_{yy'}y) - (\phi_{yx}+\phi_{yy}y'+\phi_{yy'}y)$$ (4.5.10)
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Putting (4.5.10) in (4.5.9),


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$$ \phi_{xy}-\phi_{yx} - \frac{d}{dx} (\phi_{xy'}-\phi_{y'x}+\phi_{yy'}y'-\phi_{y'y}y') =0 $$ (4.5.11)
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For the terms to cancel out we should have the following relations,


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$$ \phi_{xy}=\phi_{yx}, \phi_{xy'}=\phi_{y'x}, \phi_{yy'}= \phi_{y'y} $$ (4.5.11)
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Hence Proved.

Author and References

 * Solved and Typed by -- Pushkar Mishra